Simplify the following expression: $y = \dfrac{-4x^2+1x+3}{-4x - 3}$
First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(-4)}{(3)} &=& -12 \\ {a} + {b} &=& &=& {1} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $-12$ and add them together. Remember, since $-12$ is negative, one of the factors must be negative. The factors that add up to ${1}$ will be your ${a}$ and ${b}$ When ${a}$ is ${-3}$ and ${b}$ is ${4}$ $ \begin{eqnarray} {ab} &=& ({-3})({4}) &=& -12 \\ {a} + {b} &=& {-3} + {4} &=& 1 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({-4}x^2 {-3}x) + ({4}x +{3}) $ Factor out the common factors: $ x(-4x - 3) - 1(-4x - 3)$ Now factor out $(-4x - 3)$ $ (-4x - 3)(x - 1)$ The original expression can therefore be written: $ \dfrac{(-4x - 3)(x - 1)}{-4x - 3}$ We are dividing by $-4x - 3$ , so $-4x - 3 \neq 0$ Therefore, $x \neq -\frac{3}{4}$ This leaves us with $x - 1; x \neq -\frac{3}{4}$.